THanks for looking at myquestion. Cheers! A catapult on a clifflaunches a large

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A catapult on a clifflaunches a large round rock towards a ship on the ocean below. Therock leaves the catapult from a height H of 32.0 m above sea level,directed at an angle above the horizontal with an unknownspeed v0.
Theprojectile remains in flight for 6.00 seconds and travels ahorizontal distance D of 142.0 m. Assuming that air friction can beneglected, calculate the value of the angle.

Calculatethe speed at which the rock is launched.
To what height above sea level does the rockrise?

A catapult on a clifflaunches a large round rock towards a ship on the ocean below. Therock leaves the catapult from a height H of 32.0 m above sea level,directed at an angle above the horizontal with an unknownspeed v0.
Theprojectile remains in flight for 6.00 seconds and travels ahorizontal distance D of 142.0 m. Assuming that air friction can beneglected, calculate the value of the angle.

Calculatethe speed at which the rock is launched.
To what height above sea level does the rockrise?
A catapult on a clifflaunches a large round rock towards a ship on the ocean below. Therock leaves the catapult from a height H of 32.0 m above sea level,directed at an angle above the horizontal with an unknownspeed v0.
Theprojectile remains in flight for 6.00 seconds and travels ahorizontal distance D of 142.0 m. Assuming that air friction can beneglected, calculate the value of the angle.

Calculatethe speed at which the rock is launched.
To what height above sea level does the rockrise?

Theprojectile remains in flight for 6.00 seconds and travels ahorizontal distance D of 142.0 m. Assuming that air friction can beneglected, calculate the value of the angle.

Calculatethe speed at which the rock is launched.
To what height above sea level does the rockrise?

Explanation / Answer

Kinematics: X = xo + vxot +(1/2)axt2 142.0m = 0m + vx(6.00s) + 0 vx = (142.0m)/6.00s = 23.7 m/s Y = yo + vyot +(1/2)ayt2 0 = 32.0m + vyo(6.00s) +(1/2)(-9.81m/s2)(6.00s)2 vyo = -32.0m/(6.00s) +(4.91m/s2)(6.00s)2/(6.00s) vyo = -5.33 m/s + 29.5 m/s vyo = 24.2 m/s Vmag =(vxo2 +vyo2)1/2 = [(23.7m/s)2+ (24.2m/s)2]1/2 = 33.9 m/s ay = ay vy = ayt + vyo at max height (Y), vy = 0 vy = 0 = ayt + vyo =(-9.81m/s2)t + 24.2 m/s = 0 t = (-24.2m/s)/(-9.81m/s2) = 2.47 s Y = (1/2)ayt2 + vyot +yo Y = (1/2)(-9.81m/s2)(2.47s)2 +(24.2m/s)(2.47s) + 32.0m Y = -29.9m + 59.8m + 32.0m Ymax = 61.9 m Y = (1/2)(-9.81m/s2)(2.47s)2 +(24.2m/s)(2.47s) + 32.0m Y = -29.9m + 59.8m + 32.0m Ymax = 61.9 m

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